What is the inverse fourier transformation of $\frac{1}{i\sinh(\omega \tau)}$

Question

I managed to do the following:

$f(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty} F(\omega)e^{i\omega t} d\omega$

$\frac{1}{i\pi}\int_{-\infty}^{\infty} \frac{e^{\omega (it+\tau)}}{e^{2\omega \tau}-1}d\omega$

$\frac{1}{i\pi}\int_{-\infty}^{\infty} \frac{e^{\omega (it+\tau)}}{(e^{\omega \tau}-1)(e^{\omega\tau}+1)}d\omega$

I tried to change the equation above into the equation below by substituting $n = e^{\omega\tau}$

$f(t) = \frac{1}{i\tau\pi}\int_{0}^{\infty}\frac{n^{\frac{it}{\tau}}}{(n-1)(n+1)} dn$

I cannot find a way to do contour integration on the equation above.

Any help would be appreciated.

Answer 1

You can use Mathematica to calculate/verify your answer.

InverseFourierTransform[(1/I/Sinh[\[Omega]*\[Tau]]) // TrigToExp, \[Omega], t, FourierParameters -> {1, -1}]

$$ \begin{aligned} \mathrm{f}(\mathrm{t})&=\frac{1}{2 \pi} \int_{-\infty}^{\infty} F(\omega) e^{i \omega t} d \omega \\ &=\frac{1}{2 \pi} \int_{-\infty}^{\infty} \frac{1}{i \sinh (\omega \tau)} e^{i \omega t} d \omega \\ &=\frac{\tanh \left(\frac{\pi t}{2 \tau }\right)}{2 \tau } \end{aligned} $$

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when you set FourierParameters -> {a, b}, corresponding InverseFourierTransform formula is:

$$ \sqrt{\frac{|b|}{(2 \pi)^{1+a}}} \int_{-\infty}^{\infty} F(\omega) e^{-i b \omega t} d \omega . $$

Answer 2

Too long for a comment

Starting from the expression by @Oleksandr Kulkov $$f(t)=\frac{1}{\pi}\int\limits_{0}^\infty \frac{\sin \omega t}{\sinh \omega \tau} d \omega=\frac1{2\pi}\Im\int_{-\infty}^\infty\frac{e^{i\omega t}}{\sinh\omega\tau}d\omega=\frac1{2\tau}\Im\int_{-\infty}^\infty\frac{e^{i\pi\frac t\tau x}}{\sinh\pi x}dx=\frac1{2\tau}\Im \,I_0(t,\tau)$$ Then consider a rectangular contour $-R\to R\to R+i \to -R+i\to -R\,\,(R\to\infty)$, with added two arches - around $z=0$ and $z=i$: $I_1$ and $I_2$ (both clockwise; to get a closed contour). There are no poles inside the contour, and integrals $R\to R+i$ and $-R+i\to-R$ tend to zero;

therefore, $$\oint\frac{e^{i\pi\frac t\tau z}}{\sinh\pi z}dz=I_0(1+e^{-\pi\frac t\tau})+I_1+I_2=0$$ $$\Rightarrow I_0(1+e^{-\pi\frac t\tau})=\pi i\,(Res_{z=0}+Res_{z=i})\frac{e^{i\pi\frac t\tau z}}{\sinh\pi z}=i(1-e^{-\pi\frac t\tau})$$ $$\Rightarrow\, I_0=i\tanh\frac{\pi t}{2\tau}\,\Rightarrow\, f(t)=\frac1{2\tau}\Im\,I_0=\frac1{2\tau}\tanh\frac{\pi t}{2\tau}$$ what coincides with the result for $f(t)$ provided by @138 Aspen